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Yakvenalex [24]
3 years ago
13

A rock is thrown upward with a velocity of 22 meters per second from the top of a 25 meter high cliff, and it misses the cliff o

n the way back down. When will the rock be 9 meters from ground level? Round your answer to two decimal places.

Physics
2 answers:
Karo-lina-s [1.5K]3 years ago
7 0

Explanation:

Below are attachments containing the solution

Mice21 [21]3 years ago
6 0

Answer:

The rock will reach 9 m from the ground at eaxactly 5.06 s after it was initially thrown upwards.

Explanation:

We will use the equations of motion for this.

u = initial velocity of the rock = 22 m/s

g = acceleration due to gravity = -9.8 m/s²

y = vertical position of the rock at a time t = 9 m

y₀ = initial height of the rock = 25 m

t = time it takes for the rock to reach height of 9 m.

(y-y₀) = ut + 0.5gt²

(9 - 25) = 22t + 0.5(-9.8)t²

- 14 = 22t - 4.9t²

4.9t² - 22t - 14 = 0

solving this quadratic equation,

t = 5.055 s or - 0.565 s

Since time cannot be negative,

t = 5.055 s = 5.06 s

Hope this Helps!!!

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When water freezes, its volume increases by 9.05% (that is, equation). What force per unit area is water capable of exerting on
PIT_PIT [208]

Answer:

1.991 × 10^(8) N/m²

Explanation:

We are told that its volume increases by 9.05%.

Thus; (ΔV/V_o) = 9.05% = 0.0905

To find the force per unit area which is also pressure, we will use bulk modulus formula;

B = Δp(V_o/ΔV)

Making Δp the subject gives;

Δp = B(ΔV/V_o)

Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²

Thus;

Δp = 2.2 × 10^(9)[0.0905]

Δp = 1.991 × 10^(8) N/m²

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A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of t
nexus9112 [7]

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency  of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

 \epsilon = 1 - \dfrac{T_C}{T_H}

 0.25 = 1 - \dfrac{T_C}{464}

 T_C= (1 - 0.25) \times 464

 T_C= 0.75 \times 464

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now,

if the efficiency is equal to 42$ = 0.42

 T_C= (1 - 0.42) \times 464

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A ball is thrown up at a speed of 20 m/s.
ioda

Given:

(Initial velocity)u=20 m/s

At the maximum height the final velocity of the ball is 0.

Also since it is a free falling object the acceleration acting on the ball is due to gravity g.

Thus a=- 9.8 m/s^2

Now consider the equation

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Where v is the final velocity which is measured in m/s

Where u is the initial velocity which is measured in m/s

a is the acceleration due to gravity measured in m/s^2

s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.

Substituting the given values in the above formula we get

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